TheoremProved

Differentiability implies continuity

Statement

If $f:(a,b)\to\mathbb{R}$ is differentiable at $x_0\in(a,b)$ , then $f$ is continuous at $x_0$ .

Why is it true?

A curve with a well-defined tangent slope at a point cannot have a jump there: differentiability is a strictly stronger requirement than continuity, and the converse fails, as the sharp corner of f(x)=|x| at x=0 shows - continuous there, but not differentiable.

Proof sketch

For $x\neq x_0$ near $x_0$ , write $f(x)-f(x_0) = \frac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)$ . As $x\to x_0$ , the first factor tends to the finite number $f'(x_0)$ (by definition of differentiability), and the second factor tends to $0$ ; the product of a quantity converging to a finite limit and a quantity converging to $0$ tends to $0$ , so $f(x)-f(x_0)\to 0$ , i.e. $f$ is continuous at $x_0$ .

Proved by

proved 1823Augustin-Louis Cauchy

Topics that use this theorem

Derivative

Related theorems

Chain rule Product rule for derivatives

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Augustin-Louis Cauchy (1823). Résumé des leçons données à l'École royale polytechnique sur le calcul infinitésimal
Walter Rudin (1976). Principles of Mathematical Analysis